Problem: Multiply the following complex numbers: $({-3i}) \cdot ({3+3i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({-3i}) \cdot ({3+3i}) = $ $ ({0} \cdot {3}) + ({0} \cdot {3}i) + ({-3}i \cdot {3}) + ({-3}i \cdot {3}i) $ Then simplify the terms: $ (0) + (0i) + (-9i) + (-9 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 0 + (0 - 9)i - 9i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 0 + (0 - 9)i - (-9) $ The result is simplified: $ (0 + 9) + (-9i) = 9-9i $